Orthogonal Splitting: how it relates to the Representation Problem of Quadratic Forms

 As previously mentioned, one of the key questions of the theory of the quadratic space is:

Given an \(R\)-quadratic module \(V\) and \(\alpha \in F = \text{Frac}(R)\), when does \(V\) represents \(\alpha\)? (Written \(\alpha \to V\)). 

1. Orthogonal Splitting

Let \(V\) be a quadratic space over a field \(F\). 

We write \(V = W_1 \perp W_2\) if there are subspaces \(W_i\) such that \(V = W_1 \oplus W_2\) and every element of \(W_1\) is orthogonal to every element of \(W_2\). 

Given a subspace \(W\) of \(V\), we define its orthogonal complement \(W^{\perp} \) to be the set of vectors in \(V\) orthogonal to every vector in \(W\). 

Question. Is it true that if  \(V = W \perp U\) then \( U = W^{\perp}\)?

Answer. No because \(W^{\perp} \cap W\) might be nonzero so the sum \( W + W^{\perp}\) might not be direct. 

 We call the set \( \text{rad}(W):= W \cap W^{\perp}\) the radical of \(W\). Thus the radical of \(W\) consists of \(w \in W\) that are orthogonal to every element of \(W\).  If \( \text{rad}(W) = \{0\} \) then \(W\) is said to be regular. Above we see that if \(W\) is NOT regular then \(V \neq W \perp W^{\perp}\). It turns out that this is the only obstruction. 

Before showing that this is the case, let us find another characterization of regularity. 

1.1. Regularity in terms of Discriminant.

 Recall every quadratic module over a field has an orthogonal basis. So suppose \(V \cong \langle \alpha_i, \ldots, \alpha_n\rangle\) in some  orthogonal basis  \(\{e_1, \ldots, e_n\} \). Recall that this means \( B(e_i, e_i) = \alpha_i\) and the Gram matrix of \(V\) with respect to \( \{e_i\}\) is therefore 

\[A = \begin{pmatrix}\alpha_1 & 0 & \cdots & 0 \\ 0 & \alpha_2 & \cdots & 0 \\ 0 & 0 & \cdots &\alpha_n \end{pmatrix}. \]

In particular \(d V \equiv \det A = \prod_i \alpha_i \).

Now for every vector \(x = \sum c_i e_i \in V\) we have \(B(x, e_i) = c_i B(e_i, e_i).\)  So \[x \in \text{rad}(V) \iff B(x, e_i) = 0 \,\,\forall \, i \iff c_i B(e_i, e_i) = c_i \alpha_i= 0 \,\,\,\, \forall \, i.\]

Thus we have \[\text{rad}(V) \neq \{0\} \iff \exists c_i \neq 0 \text{ s.t } c_i \alpha_i = 0 \iff \text{ one of the } \alpha_i \text{ is } 0 \iff \det A = 0. \]

Thus we have established the following.

Prop. A quadratic \(F\)-space \(V\) is regular \(\iff\) the determinant of its Gram matrix is nonzero.

Remark: We will often abuse notation and write the second condition as \(dV \neq 0\) though strictly speaking \(dV \) is the equivalence class of \(\det A\) in \(F^*/(F^*)^2\). 

 

1.2. Regular Subspaces Split. 

For regular subspaces, Orthogonal Complements are truly Complements!

Theorem. Let \(W\) be a regular subspace of \(V\). Then we have  

•  \(V = W \perp W^{\perp}\). 
• Furthermore, if \(V = W \perp U\) then \(U = W^{\perp}.\) 

Proof of Theorem. 

• From the above discussion, we already see that \(W \cap W^{\perp} = \{0\} \).  It remains to show that \(W + W^{\perp} = V\). Let \(x \in V\), we want to show that there exists \(w \in W\) such that \((x - w) \in W^{\perp}\), i.e. \(B(x - w, - )|_W \equiv 0\). Or equivalently \(B(x,-) = B(w, -)\) on \(W\). 

Since \(B(x, -)|_W \in W^*\), it suffices for us to show that every linear functional in \(W^*\) is of the form \(B(w, -)\) for some \(w \in W\). 

Prop. Define a map \(f: W \to W^*\) by \(w \mapsto B(w, -)\). Then \(f\) is an isomorphism \( \iff W \) is regular. 

proof of the proposition. Fix a basis \(\{e_1, \ldots, e_n\}\) of \(W\) and let \(A = (a_{ij}) = \left( B(e_i, e_j)\right)\) be the corresponding Gram matrix of \(W\). Then we claim that the matrix of \(f\) with respect to the bases \( \{e_i\}, \{e_i^*\}\) is \(A\) 

. 

Indeed, let's compute \(f(e_i) \in W^*\). We have \(f(e_i)(e_j) = B(e_i, e_j) = a_{ij}\), so \(f(e_i) = \sum_j a_{ij} e_j^*\). 

Thus \(f\) is an isomorphism iff \(A\) is invertible. The later holds iff \( \det A \neq 0\), and by the previous section on regularity, that is equivalent to \(W\) being regular. \(\square\)

•  Now, finally suppose \(V = W \perp U \) then \(U \subset W^{\perp}\) and \( \dim U = \dim V - \dim W = \dim W^{\perp}\). Therefore \(U = W^{\perp}\). \(\blacksquare\)

2. Relation to Representations.

Corollary. Let \( \alpha \in F^*\). Then \[ \alpha \to V \iff \text{ there is a splitting } V = \langle \alpha \rangle \perp U .\]

Proof. Suppose \( q(w) = \alpha\) for some \(w \in V\). Let \(W = \text{span}_F(v) \subset V\). Then \(W\) is clearly regular: a vector \((c w) \in W \) lies in its radical \(\iff\) \(c \alpha = B(cw, w) = 0\), i.e. \(c = 0\).  Thus by the Theorem in Section 1.2, we must have an orthogonal splitting \(V = W \perp W^{\perp} = \langle \alpha\rangle \perp W^{\perp}.\) 

Conversely suppose \(V = \langle \alpha \rangle \perp U \). This means in particular that \(V\) contains a subspace \( W \cong \langle \alpha \rangle\) in some basis \(\{w\}\) of \(W\). Thus \(q(w) = \alpha\). \(\blacksquare\)